Edexcel Chemistry A2 Unit 5 ~ Wednesday 19th June 2013 (Now Closed)

Original post by AS01

No we dont :smile:

Oh thank god for that haha. Thanks. :smile:

Reply 542

AS01

12

Original post by LeaX

Oh thank god for that haha. Thanks. :smile:

remembering that would be daunting lol!

Reply 543

Yeah but after using some of them so often, you can probably remember that Glycine is the simplest one, with a H as the R group in it.

Although that's the only one I know thus far XD
Won't bother learning them either really.

Reply 544

BorisGlichington

13

Original post by AS01

I applied to St.George's, Queen Mary, Sheffield and Birmingham. Did you get offers?

Woo! Another medic!

Reply 545

David Tennant

OP

Original post by AS01

I applied to St.George's, Queen Mary, Sheffield and Birmingham. Did you get offers?

Yeah, Got an offer from Sheffield...Not sure how though!

Reply 546

Original post by David Tennant

Yeah, Got an offer from Sheffield...Not sure how though!

Congratulations :smile:

Ummm with Vanadium's oxidation states, how do you know which oxidising agent to use to oxidise it from +2 to +4 (as an example) or reduce it from +5 to +3 etc

and do we have to know the equations for the oxidation/reduction reactions of vanadium? I know we do for Copper/Chromium compounds

Reply 547

Original post by Mollymod

Congratulations :smile:

Ummm with Vanadium's oxidation states, how do you know which oxidising agent to use to oxidise it from +2 to +4 (as an example) or reduce it from +5 to +3 etc

and do we have to know the equations for the oxidation/reduction reactions of vanadium? I know we do for Copper/Chromium compounds

Wait so you can't use the oxidisting agent... potassium manganate (VII) all the time ??? :redface:

and what equations do we need to remember :redface:

Also a quick question - In the Facer book... if I come across anything to do with DNAs and blood, I can just skip it right ? :tongue:

Thanks

Reply 548

Original post by posthumus

Wait so you can't use the oxidisting agent... potassium manganate (VII) all the time ??? :redface:

and what equations do we need to remember :redface:

Also a quick question - In the Facer book... if I come across anything to do with DNAs and blood, I can just skip it right ? :tongue:

Thanks

What do you mean? You have to know that blood needs to maintain a pH so there's a buffer involved with pH regulation in blood.

DNA, I don't really remember much coming up about DNA to do with Chemistry. There's a bit about protein synthesis but I dont know what you're referring to here.

Nope, you have to use zinc for some of them, and Nitric acid for others; I dont know off the top of my head. Chemguide has the stuff somewhere.

And like [Cu(H2O)6]2+(aq) +2OH-(aq) -> [Cu(H2O)4(OH)2] (s)

Reply 549

Original post by Mollymod

What do you mean? You have to know that blood needs to maintain a pH so there's a buffer involved with pH regulation in blood.

DNA, I don't really remember much coming up about DNA to do with Chemistry. There's a bit about protein synthesis but I dont know what you're referring to here.

Nope, you have to use zinc for some of them, and Nitric acid for others; I dont know off the top of my head. Chemguide has the stuff somewhere.

And like [Cu(H2O)6]2+(aq) +2OH-(aq) -> [Cu(H2O)4(OH)2] (s)

I am utterly clueless lol :frown:

Thank-you for the reply

Reply 550

Original post by Mollymod

What do you mean? You have to know that blood needs to maintain a pH so there's a buffer involved with pH regulation in blood.

DNA, I don't really remember much coming up about DNA to do with Chemistry. There's a bit about protein synthesis but I dont know what you're referring to here.

Nope, you have to use zinc for some of them, and Nitric acid for others; I dont know off the top of my head. Chemguide has the stuff somewhere.

And like [Cu(H2O)6]2+(aq) +2OH-(aq) -> [Cu(H2O)4(OH)2] (s)

vo2^+ +2h+ + -> vo^2+ + h20

vo^2+ + 2h+ -> v3+ h20

basically i think you look at the different e values in the data booklet .... but i think Zn can reduce vanadium from 5+ all the way to 2+ === this is according to chem guide

http://www.chemguide.co.uk/inorganic/transition/vanadium.html

Reply 551

Original post by jojo1995

vo2^+ +2h+ + -> vo^2+ + h20

vo^2+ + 2h+ -> v3+ h20

basically i think you look at the different e values in the data booklet .... but i think Zn can reduce vanadium from 5+ all the way to 2+ === this is according to chem guide

http://www.chemguide.co.uk/inorganic/transition/vanadium.html

Would you be able to explain why that is ? :frown:

Zinc has full orbitals :frown:

Reply 552

Original post by jojo1995

vo2^+ +2h+ + -> vo^2+ + h20

vo^2+ + 2h+ -> v3+ h20

basically i think you look at the different e values in the data booklet .... but i think Zn can reduce vanadium from 5+ all the way to 2+ === this is according to chem guide

http://www.chemguide.co.uk/inorganic/transition/vanadium.html

So if the E value is higher than the E cell of the Vanadium half equation, it can reduce it?

I mean, let's say

VO3- +5e + 6H+ --------> 3H2O + V E Cell's 1.1V
and Zn -----> Zn2+ + 2e E Cell's 1.12V

would that mean the Zn can reduce the VO3- because it has a bigger E Value or am I completely mistaken?

Reply 553

Original post by Mollymod

So if the E value is higher than the E cell of the Vanadium half equation, it can reduce it?

I mean, let's say

VO3- +5e + 6H+ --------> 3H2O + V E Cell's 1.1V
and Zn -----> Zn2+ + 2e E Cell's 1.12V

would that mean the Zn can reduce the VO3- because it has a bigger E Value or am I completely mistaken?

That should be right ? If it is feasible (that way round) meaning the electrode potential is positive, then that means the reaction would happen.

Since VO3- is getting reduced and Zinc oxidized, that makes zinc the reducing agent ? :smile:

EDIT: And are you sure both the Electrode potentials will be positive, or did you make those values up ? :tongue:

(edited 11 years ago)

Reply 554

Original post by posthumus

Would you be able to explain why that is ? :frown:

Zinc has full orbitals :frown:

Doc4.odt

in the attatchment you can see that the e potential for the reduction of vanadium is more positive than that of zn - so the reaction between the vanadium and zinc will occur

zinc will lose its 4s electron so the 3d electrons arent affected

Reply 555

Original post by Mollymod

So if the E value is higher than the E cell of the Vanadium half equation, it can reduce it?

I mean, let's say

VO3- +5e + 6H+ --------> 3H2O + V E Cell's 1.1V
and Zn -----> Zn2+ + 2e E Cell's 1.12V

would that mean the Zn can reduce the VO3- because it has a bigger E Value or am I completely mistaken?

isnt the e value for the reduction of zn2+ -> zn is -0.76 ?

(edited 11 years ago)

Reply 556

Original post by jojo1995

Doc4.odt

in the attatchment you can see that the e potential for the reduction of vanadium is more positive than that of zn - so the reaction between the vanadium and zinc will occur

zinc will lose its 4s electron so the 3d electrons arent affected

If its lost only its 4s electrons then how would Vanadium go from +5 state to +2 ? :frown:

and the 4s is full anyway ? :frown:

so wouldn't it be stable anyway.

Thanks for that file :smile:

but How do I work out the Electrode cell potential? First I multiply the top by 2 right ? & then I switch the bottom one (change -0.76 to +0.76)

So when adding the 2 up I get Ecell = 2.76 V ?

Reply 557

Original post by jojo1995

isnt the e value for the reduction of zn2+ -> zn is -0.76 ?

I think she made them up :tongue:

the second equation would be positive because her equation is the other way around...

Also sorry I'm like whining today :tongue:

because I really don't understand anything... and I have so much to learn !

Reply 558